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Circular motion help

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s.

    Part A:Find the speed of the passengers when the Ferris wheel is rotating at this rate.
    -I used V=(2piR)/T and got 5.24 which was correct.

    Part B:A passenger weighs 862 N at the weight-guessing booth on the ground. What is his apparent weight at the lowest point on the Ferris wheel?
    -This is where I am completely thrown off course. Is it asking for the normal force? I thought apparent weight was just m*g?

    Part C:What is his apparent weight at the highest point on the Ferris wheel?
    -Same issue with Part B

    Part D:What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?
    -I would need the proper formula from Part B to find Part D, so I'm stuck here too.

    Part E:What then would be the passenger's apparent weight at the lowest point?
    -Same issue as Part D


    2. Relevant equations

    V=(2piR)/T
    Sum of the forces in the y direction= N=m(g-(v^2/R))

    3. The attempt at a solution

    I attempted to use that formula but it said it was wrong so I have no clue what to do now. :frown:
     
  2. jcsd
  3. Sep 28, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, the apparent weight equals the normal force.
    No, m*g is the true weight.
     
  4. Sep 28, 2008 #3
    Hi.

    what do you know about the different between ferris wheel and a loop or circle.
    Here is what i know At the lowest point in frerris wheel the
    Normal force = (mv^2)/2 + mg
    And At the highest point would be
    Normal force = mg - (mv^2)/2

    weight = mg ............
     
  5. Sep 28, 2008 #4
    Doc Al you are the man! That little answer helped me so much thanks!
     
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